A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solu- tion in the calorimeter (q 5 23.50 kJ), resulting in a temperature rise of 7.32 8C. In a subsequent experiment, 100.0 mL of 0.200 m HBr(aq) and 100.0 mL of 0.200 m KOH(aq) were mixed in the same calorimeter and the temperature rose by 2.49 8C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

I keep getting +1.19kJ instead of -1.19kJ. Could someone please explain how to get -1.19kJ?

try to think of the - and + signs as how we express the energy change from one perspective. Like it was said before, the reaction is exothermic so it'll leave to it's surroundings, lessening the amount of heat available. From this perspective, heat is being taken away and moved somewhere else.